Misc 15 - Show altitude of cone of maximum volume inscribed - Miscellaneous

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  1. Chapter 6 Class 12 Application of Derivatives
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Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is ﷐4𝑟﷮3﷯ . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let ∠ BOC = θ Now, AC = AO + OC H = r + r cos θ H = r (1 + cos θ) And, R = r sin θ We need to maximize volume of cone. Volume of the cone is V = ﷐1﷮3﷯𝜋﷐𝑅﷮2﷯𝐻 V = ﷐1﷮3﷯𝜋﷐𝑟﷮2﷯﷐﷐sin﷮2﷯﷮θ﷯ r (1 + cos θ) V = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯﷐﷐sin﷮2﷯﷮θ﷯ (1 + cos θ) Finding ﷐𝑑𝑣﷮𝑑𝜃﷯ ﷐𝑑𝑣﷮𝑑𝜃﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯ ﷐2﷐sin﷮θ﷐cos﷮θ (1+﷐cos﷮θ)+﷐﷐sin﷮2﷯﷮θ(−﷐sin﷮θ)﷯﷯﷯﷯﷯﷯ ﷐𝑑𝑣﷮𝑑𝜃﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯ ﷐2﷐sin﷮θ﷐cos﷮θ (1+﷐cos﷮θ)−﷐﷐sin﷮3﷯﷮θ﷯﷯﷯﷯﷯ ﷐𝑑𝑣﷮𝑑𝜃﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯ ﷐2﷐sin﷮θ﷐cos﷮θ (1+﷐cos﷮θ)−﷐﷐sin﷮3﷯﷮θ﷯﷯﷯﷯﷯ ﷐𝑑𝑣﷮𝑑𝜃﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯ sin θ ﷐2﷐cos﷮θ+﷐2 ﷐cos﷮2﷯﷮θ −﷐﷐sin﷮2﷯﷮θ﷯﷯﷯﷯ ﷐𝑑𝑣﷮𝑑𝜃﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯ sin θ ﷐2﷐cos﷮θ+﷐2 ﷐cos﷮2﷯﷮θ−(1−﷐﷐cos﷮2﷯﷮θ﷯﷯﷯)﷯ ﷐𝑑𝑣﷮𝑑𝜃﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯ sin θ ﷐3﷐﷐cos﷮2﷯﷮θ+﷐2 𝑐𝑜𝑠﷮θ−1﷯﷯﷯ ﷐𝑑𝑣﷮𝑑𝜃﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯ sin θ ﷐3﷐﷐cos﷮2﷯﷮θ+﷐3 𝑐𝑜𝑠﷮𝜃﷯﷯ −﷐cos﷮𝜃﷯ −1﷯ ﷐𝑑𝑣﷮𝑑𝜃﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯ sin θ ﷐3﷐𝑐𝑜𝑠 ﷮θ﷯ ﷐﷐cos﷮𝜃﷯+1﷯−1﷐﷐cos﷮𝜃﷯+1﷯﷯ ﷐𝑑𝑣﷮𝑑𝜃﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯ sin θ ﷐3﷐𝑐𝑜𝑠﷮θ−1﷯﷯﷐﷐𝑐𝑜𝑠﷮θ+1﷯﷯ Putting ﷐𝑑𝑣﷮𝑑𝜃﷯ = 0 ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯ sin θ ﷐﷐𝑐𝑜𝑠﷮θ+1﷯﷯ ﷐3﷐𝑐𝑜𝑠﷮θ−1﷯﷯ = 0 sin θ (cos θ + 1) (3 cos θ − 1) = 0 So, ﷐𝑐𝑜𝑠﷮θ﷯= ﷐1﷮3﷯ Thus, H = r (1 + cos θ) H = r ﷐1 + ﷐1﷮3﷯﷯ H = ﷐4𝑟﷮3﷯ This value is either maxima or minima. So, finding ﷐﷐𝑑﷮2﷯𝑉﷮𝑑﷐𝜃﷮2﷯﷯ ﷐𝑑𝑉﷮𝑑𝜃﷯ = ﷐﷐1﷮3﷯𝜋﷐𝑟﷮3﷯﷐sin﷮𝜃﷯﷐3 cos2 𝜃 + 2﷐cos﷮𝜃﷯− 1﷯﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯﷐﷐cos﷮𝜃﷯﷐3﷐﷐cos﷮2﷯﷮𝜃﷯+2﷐cos﷮𝜃﷯−1﷯+﷐sin﷮𝜃﷯﷐6﷐cos﷮𝜃﷯﷐−𝑠𝑖𝑛﷯−2﷐sin﷮𝜃﷯﷯﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯﷐3﷐﷐cos﷮3﷯﷮𝜃﷯+2﷐﷐cos﷮2﷯﷮𝜃﷯−﷐cos﷮𝜃﷯+﷐sin﷮𝜃﷯﷐−6﷐cos﷮𝜃﷯﷐sin﷮𝜃﷯−2﷐sin﷮𝜃﷯﷯﷯ = ﷐1﷮3﷯𝜋﷐𝑟﷮3﷯﷐3﷐﷐cos﷮2﷯﷮𝜃﷯+2﷐﷐cos﷮2﷯﷮𝜃﷯−﷐cos﷮𝜃﷯−6﷐﷐sin﷮2﷯﷮𝜃﷯﷐cos﷮𝜃﷯−2﷐﷐sin﷮2﷯﷮𝜃﷯﷯ Now, ﷐cos﷮𝜃﷯=﷐1﷮3﷯ And sin2 θ = 1 − cos2 θ = 1 − ﷐﷐﷐1﷮3﷯﷯﷮2﷯= 1 − ﷐1﷮9﷯ = ﷐8﷮9﷯ Putting values in ﷐﷐𝑑﷮2﷯𝑉﷮𝑑﷐𝜃﷮2﷯﷯ ﷐﷐𝑑﷮2﷯𝑉﷮𝑑﷐𝜃﷮2﷯﷯ = ﷐1﷮3﷯ 𝜋﷐𝑟﷮3﷯ ﷐3﷐﷐﷐1﷮3﷯﷯﷮3﷯+2﷐﷐﷐1﷮3﷯﷯﷮2﷯−﷐1﷮3﷯−6﷐﷐1﷮3﷯﷯﷐﷐8﷮9﷯﷯−2﷐﷐8﷮9﷯﷯﷯ = ﷐1﷮3﷯ 𝜋﷐𝑟﷮3﷯ ﷐3﷐﷐1﷮27﷯﷯+2﷐﷐1﷮9﷯﷯−﷐1﷮3﷯−6﷐﷐8﷮27﷯﷯−2﷐﷐8﷮9﷯﷯﷯ = ﷐1﷮3﷯ 𝜋﷐𝑟﷮3﷯ ﷐﷐1﷮9﷯+ ﷐2﷮9﷯−﷐1﷮3﷯−﷐16﷮9﷯−﷐16﷮9﷯﷯ = ﷐1﷮3﷯ 𝜋﷐𝑟﷮3﷯ ﷐﷐−32﷮9﷯﷯ = ﷐−32𝜋﷐𝑟﷮3﷯﷮27﷯ Thus, ﷐﷐𝑑﷮2﷯𝑣﷮𝑑﷐𝜃﷮2﷯﷯ < 0 H = ﷐𝟒𝒓﷮𝟑﷯ Hence proved

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